Lue Vang Mr. Kellogg AP Physics - Pd. 7 5 February 2012 Performed On: 01 February 2012
Elasticity and Inelasticity Lab
Purpose: This lab will study elastic and inelastic collisions, specifically their effect on a system’s momentum.
Background: During collisions, momentum is conserved, but KE may change, thus resulting in two types of collision: elastic and inelastic. In elastic collisions, bother momentum and KE are conversed while in inelastic collisions, momentum is conserved but KE is not.
In the case of completely inelastic collisions, a moving object collides with a stationary one and the two stick together as they move in at the same velocity. For such instances,
P1 = P2 m1v1 = (m1+m2)v
Kf/Ki = m1/(m1 + m2)
For elastic collisions, the momentums and/or kinetic energy before and after collisions are equal.
P1 = P2 m1v1 = m2v2
KE1 = KE2 1/2m1v12 = 1/2m2v22
There are special cases properties when a moving object collides with a stationary one. When each one’s masses are the same (m1 = m2), then m1 stops completely and the velocity of m2 after the collision is equal to the initial velocity of m1. If the moving mass m1 is massively larger than the stationary m2, then the velocity of m1 stays close to its initial and m2’s is about two times that of m1’s original velocity. In the opposite case, m1’s velocity is equal and opposite while m2 stays stationary.
m1 = m2 … then v1 = 0 and v2 = v1o
m1 >> m2 … v1 ≈ v1o ... v2 ≈ 2v1o
m1 << m2 … v1 ≈ -v1o ... v2 ≈ 0
Hypothesis: For Elastic I, both conservation of momentum and energy should be applied, and since both objects are of equal mass, m1 should stop after the collision, and m2 should start traveling at the initial velocity of m1.
For Inelastic I, only conservation of momentum will apply. Since it’s inelastic, the two object will likely travel together at the same velocity after collision, but the velocity will be significantly slower than the original velocity of m1.
For Elastic II, both conservation of momentum and energy should be applied, but since m1 is about two times the size of m2, then v1 should continue to travel at approximately the same velocity after the collision while m2 will move at about two times the velocity of m1.
For Inelastic II, only conservation of momentum will apply. Since it’s inelastic, the two object will likely travel together at the same velocity after collision, but the velocity will be significantly slower than the original velocity of m1.
Apparatus:
1 Functional Computer/Laptop
Lab Video Files
Logger Pro Program
Procedure: Open the videos in logger pro. Then use the video analysis tool to track the positions of the blue and red cup for each video. Finally, find each cup’s velocity for each video by graphing the data points of position(x) over time (s). After finding the velocities, use them to calculate momentum and kinetic energy for each system.
Data:
Graph 1.0 - Elastic I - Cart 1 m1 = 0.515 kg v1o = 0.698 m/s v1f = 0.042 m/s
Graph 1.1 - Elastic I - Cart 2 m2= 0.515 kg v2o = 0.000 m/s v2f = 0.576 m/s
Graph 2.0 - Inelastic I - Cart 1 m1 = 0.515 kg v1o = 0.576 m/s v1f = 0.192 m/s
Graph 2.1 - Inelastic I - Cart 2 m2= 0.515 kg v2o = 0.000 m//s v2f = 0.194 m/s
Graph 3.0 - Elastic II - Cart 1 m1 = 1.015 kg v1o = 0.664 m/s v1f = 0.324 m/s
Graph 3.1 - Elastic II - Cart 2 m2= 0.551kg v2o = 0.000 m/s v2f = 0.5932 m/s
Graph 4.0 - Inelastic II - Cart 1 m1 = 1.015 kg v1o = 0.561 m/s v1f = 0.311 m/s
Graph 4.1 - Inelastic II - Cart 2 m2= 0.551kg v2o = 0.000 m/s v2f = 0.308 m/s
Analysis: For Equations, refer back to the background section.
Conclusion: The video analysis shows proves the hypothesis to be generally correct.
In Elastic I, it can be seen that almost all momentum and KE are transferred to cart 2 (0.297kg*m/s and 0.085W) from cart 1 (0.359kg*m/s and 0.125W). They’re relatively equal, but cart 2’s values are less due some energy is lost due to friction and other such conveniences. Also, notice the velocity for cart 2 after the collision (0.576m/s) is relatively equal to that of cart 1 (0.698m/s). And after the collision, cart 1’s velocity is also approximately zero (0.042m/s). For our purposes, this system is pretty conservative. In analyzing the conservation charts, its conservation results are relatively high (89% and 68%) compared to the other trials.
In Inelastic I, it can be seen that momentum is relatively conserved. It’s over 100% because the tools used to analyze the video aren’t very accurate. When looking at energy, kinetic energy is almost completely drained. In fact, only about 22% of the original amount is left. This proves the inelastic property of this trial. Also, as hypothesized, the two carts traveled together after the collision and, afterwards, traveled at the same velocity of about 0.193m/s.
Elastic II, like Elastic I, shows the system to be relatively conservative in both momentum with 97% conservation and kinetic energy with 67% conservation. Similar to Elastic I, a lot of the energy from cart 1 is transferred to cart 2, and as hypothesized, cart 2’s velocity (0.593 m/s) becomes about two times that of cart 1’s 0.324m/s. Unlike Elastic I though, cart 1 doesn’t transfer as much momentum and/or kinetic energy to the cart 2.
In inelastic II, the carts can be seen sticking together again. And like inelastic I, both carts traveled together at relatively the same velocity (0.309m/s) after collision. Its momentum is conserved. In fact the conserved momentum is over 100%, which again is due to inaccurate data-gathering equipment and scientist(s). Nonetheless, the data proves the point because the conservation of kinetic energy is significantly low at 47%.
Mr. Kellogg
AP Physics - Pd. 7
5 February 2012
Performed On: 01 February 2012
Elasticity and Inelasticity Lab
Purpose:
This lab will study elastic and inelastic collisions, specifically their effect on a system’s momentum.
Background:
During collisions, momentum is conserved, but KE may change, thus resulting in two types of collision: elastic and inelastic. In elastic collisions, bother momentum and KE are conversed while in inelastic collisions, momentum is conserved but KE is not.
In the case of completely inelastic collisions, a moving object collides with a stationary one and the two stick together as they move in at the same velocity. For such instances,
P1 = P2
m1v1 = (m1+m2)v
Kf/Ki = m1/(m1 + m2)
For elastic collisions, the momentums and/or kinetic energy before and after collisions are equal.
P1 = P2
m1v1 = m2v2
KE1 = KE2
1/2m1v12 = 1/2m2v22
There are special cases properties when a moving object collides with a stationary one. When each one’s masses are the same (m1 = m2), then m1 stops completely and the velocity of m2 after the collision is equal to the initial velocity of m1. If the moving mass m1 is massively larger than the stationary m2, then the velocity of m1 stays close to its initial and m2’s is about two times that of m1’s original velocity. In the opposite case, m1’s velocity is equal and opposite while m2 stays stationary.
m1 = m2 … then v1 = 0 and v2 = v1o
m1 >> m2 … v1 ≈ v1o ... v2 ≈ 2v1o
m1 << m2 … v1 ≈ -v1o ... v2 ≈ 0
Hypothesis:
For Elastic I, both conservation of momentum and energy should be applied, and since both objects are of equal mass, m1 should stop after the collision, and m2 should start traveling at the initial velocity of m1.
For Inelastic I, only conservation of momentum will apply. Since it’s inelastic, the two object will likely travel together at the same velocity after collision, but the velocity will be significantly slower than the original velocity of m1.
For Elastic II, both conservation of momentum and energy should be applied, but since m1 is about two times the size of m2, then v1 should continue to travel at approximately the same velocity after the collision while m2 will move at about two times the velocity of m1.
For Inelastic II, only conservation of momentum will apply. Since it’s inelastic, the two object will likely travel together at the same velocity after collision, but the velocity will be significantly slower than the original velocity of m1.
Apparatus:
Procedure:
Open the videos in logger pro. Then use the video analysis tool to track the positions of the blue and red cup for each video. Finally, find each cup’s velocity for each video by graphing the data points of position(x) over time (s).
After finding the velocities, use them to calculate momentum and kinetic energy for each system.
Data:
Graph 1.0 - Elastic I - Cart 1
m1 = 0.515 kg
v1o = 0.698 m/s
v1f = 0.042 m/s
Graph 1.1 - Elastic I - Cart 2
m2= 0.515 kg
v2o = 0.000 m/s
v2f = 0.576 m/s
Graph 2.0 - Inelastic I - Cart 1
m1 = 0.515 kg
v1o = 0.576 m/s
v1f = 0.192 m/s
Graph 2.1 - Inelastic I - Cart 2
m2= 0.515 kg
v2o = 0.000 m//s
v2f = 0.194 m/s
Graph 3.0 - Elastic II - Cart 1
m1 = 1.015 kg
v1o = 0.664 m/s
v1f = 0.324 m/s
Graph 3.1 - Elastic II - Cart 2
m2= 0.551kg
v2o = 0.000 m/s
v2f = 0.5932 m/s
Graph 4.0 - Inelastic II - Cart 1
m1 = 1.015 kg
v1o = 0.561 m/s
v1f = 0.311 m/s
Graph 4.1 - Inelastic II - Cart 2
m2= 0.551kg
v2o = 0.000 m/s
v2f = 0.308 m/s
Analysis:
For Equations, refer back to the background section.
Conserved Momentum = Pf(total)/ Pi(total) * 100
= (p1f + p2f)/(p1i + p2i)*100
Conserved Kinetic Energy = KEf(total)/ KEi(total) * 100
= (KE1f + KE2f)/(KE1i + KE2i)*100
Conclusion:
The video analysis shows proves the hypothesis to be generally correct.
In Elastic I, it can be seen that almost all momentum and KE are transferred to cart 2 (0.297kg*m/s and 0.085W) from cart 1 (0.359kg*m/s and 0.125W). They’re relatively equal, but cart 2’s values are less due some energy is lost due to friction and other such conveniences. Also, notice the velocity for cart 2 after the collision (0.576m/s) is relatively equal to that of cart 1 (0.698m/s). And after the collision, cart 1’s velocity is also approximately zero (0.042m/s). For our purposes, this system is pretty conservative. In analyzing the conservation charts, its conservation results are relatively high (89% and 68%) compared to the other trials.
In Inelastic I, it can be seen that momentum is relatively conserved. It’s over 100% because the tools used to analyze the video aren’t very accurate. When looking at energy, kinetic energy is almost completely drained. In fact, only about 22% of the original amount is left. This proves the inelastic property of this trial. Also, as hypothesized, the two carts traveled together after the collision and, afterwards, traveled at the same velocity of about 0.193m/s.
Elastic II, like Elastic I, shows the system to be relatively conservative in both momentum with 97% conservation and kinetic energy with 67% conservation. Similar to Elastic I, a lot of the energy from cart 1 is transferred to cart 2, and as hypothesized, cart 2’s velocity (0.593 m/s) becomes about two times that of cart 1’s 0.324m/s. Unlike Elastic I though, cart 1 doesn’t transfer as much momentum and/or kinetic energy to the cart 2.
In inelastic II, the carts can be seen sticking together again. And like inelastic I, both carts traveled together at relatively the same velocity (0.309m/s) after collision. Its momentum is conserved. In fact the conserved momentum is over 100%, which again is due to inaccurate data-gathering equipment and scientist(s). Nonetheless, the data proves the point because the conservation of kinetic energy is significantly low at 47%.